Question

Evaluate the integral, if it exists.

\(\int {\frac{{{{\csc }^2}x}}{{1 + \cot x}}} dx\)

Short answer

The integral\(\int {\frac{{{{\csc }^2}x}}{{1 + \cot x}}} dx\)exists and the value for the integral is obtained as\(I = - \ln (1 + \cot x) + c\).

Step-by-step solution

Integral definition

In mathematics, an integral is a numerical number equal to the area under a function's graph for some interval or a new function whose derivative equals the original function.

Evaluation of the integral

Evaluate the integral –

\(\int {\frac{{{{\csc }^2}x}}{{1 + \cot x}}} dx\)

Now, consider that –

\(\begin{aligned}{c}u &= 1 + \cot x\\du &= - \left( {{{\csc }^2}x} \right)dx\\dx &= - \frac{{du}}{{{{\csc }^2}x}}\end{aligned}\)

Substitute out the \(x\) and \(dx\) for \(u\) and \(du\) -

\(I = - \int {\frac{1}{u}du} \)

Substitution and further calculation

It is known that \(\int {\frac{1}{x}dx = \ln x + c} \), so –

\(I = - \ln u + c\)

Substitute back \(u = 1 + \cot x\)-

\(I = - \ln (1 + \cot x) + c\)

Therefore, the value of the integral is \(I = - \ln (1 + \cot x) + c\).